It’s me again lol you’re questions are simple the formula is given in the triangle
The answer is C 39 hz
Please look at the attached image below for the explanation
Hi there!
Acceleration = change in velocity / change in time = Δv/Δt
Thus:
a = (75 - 15)/4 = 60/4 = 15 mi/hr²
Answer:
0.166 rad/s
Explanation:
See attachment for calculations
Answer:
Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.
Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.
Explanation:
Part a
When Road is Level
The stopping sight distance is given as
Here
- SSD is the stopping sight distance which is to be calculated.
- u is the speed which is given as 60 mi/hr
- t is the perception-reaction time given as 2.5 sec.
- a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
- G is the grade of the road, which is this case is 0 as the road is level
Substituting values
So the minimum stopping sight distance is 563.36 ft.
Part b
When Road has a maximum grade of 4%
The stopping sight distance is given as
Here
- SSD is the stopping sight distance which is to be calculated.
- u is the speed which is given as 60 mi/hr
- t is the perception-reaction time given as 2.5 sec.
- a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
- G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade
For upgrade of 4%, Substituting values
<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>
For downgrade of 4%, Substituting values
<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>
As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.
Answer:
Explanation:
a) Linear density is greater on the left side because the velocity of wave in a string is inversely proportional to the linear mass density of the string
b) We should start pulse from the left hand side so the reflected wave does not get inverted because the wave traveling from the denser to lighter medium gets reflected in the same phase.
